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Homogeneous

The solution of homogeneous differential equations including the use of the D operator
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Definition

The equation Py' = Q is said to be homogeneous if P and Q are homogeneous functions of x and y of the same degree.
For example :

If \displaystyle \frac{dy}{dx} = f\left(\frac{y}{x} \right)

We can test to see whether this first order equation is homogeneous by substituting \displaystyle y = v\,x . If the result is in the form f(v) i.e. all the x's are canceled then the test is satisfied and the equation is Homogeneous.

Example:

Example - Simple example
Problem

\frac{dy}{dx} = \frac{x^2 + y^2}{2\,x^2}
Workings
Becomes
\frac{dy}{dx} = \frac{1 + v^2}{2}
Solution
There are no terms in x on the right hand side and the equation is Homogereous.

So the original equation is not homogeneous.

Methods Of Solution

A solution can be found by putting y = vx on both sides of the equation:
Example:

Example - Rational Example
Problem
\frac{dy}{dx} = \frac{x^2 + y^2}{2x^2}
Workings
Putting y - vx

Since y is a function of x so is v

\frac{dy}{dx} = v + x\,\frac{dv}{dx}
Therefore
v + x\,\frac{dv}{dx} = \frac{x^2 + y^2}{2x^2} = \frac{1 + v^2}{2}
Therefore
2x\,dv = (1 + v^2 - 2v)\,dx

Separating the variables

\frac{2\,dv}{(v - 1)^2} = \frac{dx}{x}

Integrating

But v = \frac{y}{x} so
Solution
Substituting equation (2) in equation (1)

2x = (x - y)(\ln x + C)

The General Form Of A Homogeneous Linear Equation

A homogeneous polynomial is a polynomial whose monomials with nonzero coefficients all have the same total degree.
For example : x^7+xy^6+y^7 is homogeneous polynomial .
p_0\,x^n\,\frac{d^{n}y}{dx^{n}} + p_1\,x^{n-1}\,\frac{d^{n-1}}{dx^{n-1}} + ....+\;p_{n-2}\,x^2\;\frac{d^2y}{dx^2}{ + p_{n-1}\,x\,\,\frac{dy}{dx}} + p_n\,y = fx

The method to solve this is to put \displaystyle\; x = e^t and the equation then reduces to a linear type with constant coefficients.
If x = e^t
Then \displaystyle \frac{dy}{dt} = x
Therefore \displaystyle \frac{dy}{dt} = \frac{dy}{dx}\;.\;\frac{dx}{dt} = x\,\frac{dy}{dx}

Also

\frac{d^2y}{dt^2} = \frac{d}{dx}\left(x\,\frac{dy}{dx} \right)\frac{dx}{dt} = x\,\left(\frac{dy}{dx} + x\,\frac{d^2y}{dx^2} \right)

Therefore \displaystyle x^2\,\frac{d^2y}{dx^2} + x\,\frac{dy}{dt}= \frac{d^2y}{dt^2}

Hence And

The Use Of The D Operator To Solve Homogeneous Equations

If \displaystyle x\;=z\;e^t and D = \frac{d}{dt}

The D operator is a linear operator defined as : D \equiv \frac{d}{dt} . For example : D(2x+1)=2
Then from equation (2)

x\;\frac{dy}{dx} = Dy

And from equation (3)

Example:

Example - A complex example
Problem
Solve the following Differential equation:

x^2\;\frac{d^2y}{dx^2} + 7x\,\frac{dy}{dx} + 9\,y = 18\,x^3
Workings
By putting x = e^t\; and using the D factor then the equation reduces to:-

D(D - 1)y + 7D\,y + 9y = 18\,e^{3t}
Or
(D^2 + 6D + 9)y = 18\,e^{3t}
Therefore
(D + 3)^2y = 18\,e^{3t}
Therefore
y = (A + B\,t)\;e^{-3t} + \frac{1}{(D + 3)^2}\times18\,e^{3t}

= (A + B\,t)\;e^{-3t} + \frac{18}{36}\,e^{3t}
Solution
Therefore
y = (A + B\,ln\,x)\;\frac{1}{x^3} + \frac{1}{2}\,x^{3}

Equations Which Can Be Reduced To The Homogeneous Form

Consider the following equation:

\frac{dy}{dx} = \frac{2x + 3y + 4}{4x + 5y - 10}

The equation is not Homogeneous due to the constant terms + 4 and - 10

However if we shift the origin to the point of intersection of the straight lines \displaystyle 2x + 3y + 4 = 0 and 4x + 5y - 10 = 0, then the constant terms in the differential equation will disappear.

Example:

Example - Rational Example
Problem
\frac{dy}{dx} = \frac{2x + 9y - 20}{6x + 2y - 10}
Workings
The lines \displaystyle 2x + 9y - 20 = 0 and 6x + 2y - 10 = 0 meet at the point (1, 2). We therefore make the following substitutions:

x = X + 1
y = Y + 2

The equation now becomes:

\frac{dY}{dX} = \frac{2(X + 1) + 9(Y + 2) - 20}{6(X + 1) + 2(y + 2) - 10} = \frac{2X + 9Y}{6X + 2Y}
Solution
This is homogeneous and can be solved by putting Y = v X. The solution is given by:
(2x - y)^2 = C(x + 2y - 5)

Exceptional Case

If the two straight lines are parallel, then there is no finite point of intersection and we proceed as follows:
Let \displaystyle \frac{dy}{dx} = \frac{3y - 4x - 2}{3y - 4x - 3} Put Z = 3y - 4x Then \displaystyle \frac{dZ}{dx} = 3\,\frac{dy}{dx} - 4

Thus the equation becomes:

\frac{1}{3}\left(\frac{dZ}{dx} + 4} \right) = \frac{Z - 2}{Z - 3}
Therefore
\frac{dZ}{dx} = \left( \frac{3Z - 6}{Z - 3} \right) - 4 = \frac{-\,Z + 6}{Z - 3}
Therefore
dx = -\,\left(\frac{Z - 3}{Z - 6} \right)\,dZ = -\,\left(\frac{Z - 6 + 3}{Z - 6} \right)\,dZ
= \left(-\,1 - \frac{3}{z - 6} \right)\,dZ
Therefore x = -\,Z - 3\,\ln\,(Z - 6) + K Thus 3\ln\,(3y - 4x - 6) = 3x - 3y + K